If we have the graph of a function \(f\), we can draw a sketch of the graph of \(f’\) by estimating the slope of the tangent to the graph of \(f\) at each \(x\) value. We then plot the points \((x,f'(x))\) in the \(xy\)-plane and connect them by a smooth curve whenever possible. This curve represents the graph of \(f’\).

When you seek to graph the derivative, it is often easiest to first identify the places where the slope of the tangent line is 0. Those are the places where you can hope the sign on the derivative may change.

Example 1

The graph of a function\(f\) is shown below. Give a rough sketch of the graph of \(f’.\)

Solution 1

The tangent to the curve is horizontal at \(x=0\) and \(x=2.\) \[\small{\begin{cases}
x<0: & \text{tangent line makes an acute angle with positive } x \text{ axis}\Rightarrow f'(x)>0\\
0<x<2: & \text{tangent line makes an obtuse angle with positive }x\text{ axis}\Rightarrow f'(x)<0\\
2<x: & \text{tangent line makes an acute angle with positive }x \text{ axis}\Rightarrow f'(x)>0
\end{cases}}\]
Specifically, the slope of the tangent line at \(x=-1\) is
\[m_{\text{tan}}=\frac{\text{rise}}{\text{run}}=\frac{3}{1}\Rightarrow f'(-1)=3.\]
The slope of the tangent line at \(x=1\) is
\[m_{\text{tan}}=\frac{\text{rise}}{\text{run}}=\frac{-1}{1}\Rightarrow f'(1)=-1.\]
At \(x=3\)
\[m_{\text{tan}}=\frac{\text{rise}}{\text{run}}=\frac{-3}{-1}\Rightarrow f'(3)=3.\] (See Figure 2).

If we connects these points: \[(-1,3),(0,0),(1,-1),(2,0),(3,3)\] we can sktech the graph of \(f’\) (Figure 3).

Example 2

The graph of a function \(f\) is shown in Figure 4. Give a rough sketch of the graph of \(f^\prime\). (The sketch does not need to be exact. Just show its main features.)

Solution 2

The graphs of the function and its derivative are shown in Figure 5. To see why the plot of $f’$ makes sense read on.

It is clear that the slope of the tangent line is zero at $x=-1, -0.2$ and $1$. So we have three points $(-1,0), (-0.2,0)$, and $(1,0)$ on the graph of $f’$.

For $x<-1$, the tangent line makes an acute angle with the positive direction of the $x$-axis, so $y’$ is positive in this interval. If $x$ is large negative, $y’$ is large positive; as $x$ approaches $-1$ from the left, $y’$ gets smaller and smaller until we reach $x=-1$, at which $y’=0$.

For $-1<x<-0.2$, the tangent line makes an obtuse angle with the positive direction of the $x$-axis, so $y’$ is negative in this interval. The steepest slope in this interval occurs at $B$ (i.e. $y’$ is the most negative at $B$), so we can say that $y’$ has a minimum there.

For $-0.2<y<1$, the tangent line makes an acute angle with the positive direction of the $x$-axis ($y’>0$). The slope first starts to increase, at $C$ has a local maximum, and then starts to decrease until it reaches zero at $x=1$.

For $1<x$, the slope is positive and gets larger and larger as $x$ increases.